Question

4 cos square pi by 20 minus 3 multiply 4 cos square 3 pi by 20 minus 3 is equal to​

Answer 1

$\textbf{To find:}$

$\text{The value of }$

$(4\,cos^2\frac{\pi}{20}-3){\times}(4\,cos^2\frac{3\pi}{20}-3)$

$\textbf{Solution:}$

$\text{Consider,}$

$(4\,cos^2\frac{\pi}{20}-3){\times}(4\,cos^2\frac{3\pi}{20}-3)$

$=(4\,cos^29^{\circ}-3){\times}(4\,cos^227^{\circ}-3)$

$\text{Multiply both numerator and denominator by cos\,9^{\circ}{\times}cos\,27^{\circ} }$

$=\dfrac{cos\,9^{\circ}(4\,cos^29^{\circ}-3){\times}cos\,27^{\circ}(4\,cos^227^{\circ}-3)}{cos\,9^{\circ}{\times}cos\,27^{\circ}}$

$=\dfrac{(4\,cos^39^{\circ}-3\,cos\,9^{\circ}){\times}(4\,cos^327^{\circ}-3\,cos\,27^{\circ})}{cos\,9^{\circ}{\times}cos\,27^{\circ}}$

$\text{Using the identity,}$

$\boxed{\bf\,cos3A=4\,cos^3A-3\,cosA}$

$=\dfrac{(cos\,3(9^{\circ})){\times}(cos\,3(27^{\circ}))}{cos\,9^{\circ}{\times}cos\,27^{\circ}}$

$=\dfrac{cos\,27^{\circ}{\times}cos\,81^{\circ}}{cos\,9^{\circ}{\times}cos\,27^{\circ}}$

$=\dfrac{cos\,81^{\circ}}{cos\,9^{\circ}}$

$=\dfrac{sin\,9^{\circ}}{cos\,9^{\circ}}$

$=tan\,9^{\circ}$

$=tan\frac{\pi}{20}$

$\therefore\bf(4\,cos^2\frac{\pi}{20}-3){\times}(4\,cos^2\frac{3\pi}{20}-3)=tan\frac{\pi}{20}$

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